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प्रश्न
Figure given below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.
Compare the distance travelled in part BC with the distance travelled in part AB.
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उत्तर
Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
= (2t - t) × vo
= vot
Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × vo
= (1/2) vo t
Therefore, distance travelled in part BC:distance travelled in part AB :: 2:1.
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संबंधित प्रश्न
If the displacement of an object is proportional to the square of time, then the object is moving with :
Clarify the difference.
Distance and displacement
A car travels a distance 100 m with constant acceleration and average velocity of 20 ms-1. The final velocity acquired by the car is 25 ms-1.
Find
(i) The initial velocity.
(ii) Acceleration of the car.
Draw displacement – time graph for the following situation:
When a body is moving with variable velocity.
How can you find the following?
Velocity from a displacement – time graph.
A motorbike running at 90 kmh−1, is slowed down to 54 kmh−1 by the application of brakes, over a distance of 40 m. If the brakes are applied with the same force, calculate
- total time in which bike comes to rest
- total distance travelled by bike.
Define displacement.
Fill in the boxes.
| S.No | First Move | Seconde Move | Distance (m) | Displacement |
| 1. | Move 4 metres east | Move 2 metres west | 6 | 2 m east |
| 2. | Move 4 metres north | Move 2 metres south | ||
| 3. | Move 2 metres east | Move 4 metres west | ||
| 4. | Move 5 metres east | Move 5 metres west | ||
| 5. | Move 5 metres south | Move 2 metres north | ||
| 6. | Move 10 metres west | Move 3 metres east |
Exercise Problems.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 m and 20 s?
Give one example where the displacement is zero but the distance travelled is not zero.
